LightOJ1213 Fantasy of a Summation

---

LightOJ1213 Fantasy of a Summation

---

标签

---

前言

• 我的csdn和博客园是同步的，欢迎来访danzh-博客园~

---

简明题意

• 给定n，k，mod，以及大小为n的数组a[]，求：
  $\sum_{i=1}^n\sum_{j=1}^n...\sum_{k=1}^n(a[i]+a[j]+...+a[k])$

---

思路

• 把和式拆开就能很快发现规律了。像下面这样拆开：
  $\sum_{i=1}^n(a_i+b_i)=\sum_{i=1}^na_i+\sum_{i=1}^nb_i$

---

注意事项

• 无

---

总结

• 无

---

AC代码

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

const int maxn = 1e5 + 10;

int mod;

int ksm(int a, int b)
{
	int ans = 1, base = a;
	while (b)
	{
		if (b & 1)
			ans = 1ll * ans * base % mod;
		b >>= 1;
		base = 1ll * base * base % mod;
	}
	return ans;
}

void solve()
{
	int t;
	scanf("%d", &t);
	for (int i = 1; i <= t; i++)
	{
		int n, k;
		long long sum = 0;
		scanf("%d%d%d", &n, &k, &mod);
		for (int i = 1; i <= n; i++)
		{
			int t;
			scanf("%d", &t);
			sum += t;
			sum %= mod;
		}

		printf("Case %d: %d\n", i, sum * k % mod * ksm(n, k - 1) % mod);
	}
}

int main()
{
	freopen("Testin.txt", "r", stdin);
	solve();
	return 0;
}